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Q25

Expert-verified
Found in: Page 176

Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279

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Write a paragraph proof.Given: $▱ABCD;\text{\hspace{0.17em}\hspace{0.17em}}▱BEDF$Prove: AECF is a $▱$.(Hint: A short proof is possible if certain auxiliary segments ate drawn.)

Two pairs of opposite sides $\stackrel{\to }{AE}\cong \stackrel{\to }{FC}$ and $\stackrel{\to }{AF}\cong \stackrel{\to }{CE}$in AECF are parallel therefore, AECF is a parallelogram.

See the step by step solution

Step 1. Relation between triangle DAE and triangle BCF

In triangles $\Delta DAE$ and $\Delta BCF$.

$\stackrel{\to }{AD}\cong \stackrel{\to }{BC\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}\therefore$ (ABCD is a rhombus)

$\angle ADC\cong \angle ABC\therefore$ (ABCD is a rhombus)

$\angle ADE\cong \angle CBF\text{\hspace{0.17em}\hspace{0.17em}}\therefore$ ( $\angle EDC&\angle ABF$ are the equal angle of parallelogram made of parallel sides two parallelograms)

ABCD is a parallelogram.

Similarly, $\angle ADC\cong \angle ABC$ ( ABCD is a rhombus),

$\angle EDC&\angle ABF$ are equal angles of parallelogram made of parallel sides.

$\angle ADE\cong \angle CBF$

$\stackrel{\to }{DE}\cong \stackrel{\to }{BF}$ $\left(\therefore BEDF\text{\hspace{0.17em}is a\hspace{0.17em}parallelogram}.\right)$

Therefore, by SAS postulate $\Delta DAE\cong \Delta BCF$.

By CPCT, $\stackrel{\to }{AE}\cong \stackrel{\to }{FC}$.

Step 2. Relation between triangle AFB and triangle CDE

In triangles $\Delta AFB&\Delta CDE$.

$\stackrel{\to }{DE}\cong \stackrel{\to }{BF}\therefore \left(BEDF\text{\hspace{0.17em}is a parallelogram}\right)$

Since, $\angle EDC&\angle ABF$ are equal angles of parallelogram made of parallel sides.

$\angle ADE\cong \angle CBF\text{\hspace{0.17em}\hspace{0.17em}}$

$\stackrel{\to }{DC}\cong \stackrel{\to }{BA}\therefore \left(BEDF\text{\hspace{0.17em}is a parallleogram}\right)$

Therefore, by SAS postulate $\Delta AFB\cong \Delta CED$.

By CPCT, $\stackrel{\to }{AF}\cong \stackrel{\to }{CE}$.

Step 3. State the conclusion

Since two pairs of opposite sides $\stackrel{\to }{AE}\cong \stackrel{\to }{FC}$ and $\stackrel{\to }{AF}\cong \stackrel{\to }{CE}$ in AECF are parallel, therefore, AECF is a parallelogram.

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